Derivative of function $f(x) = \sqrt{2x}+ \sqrt{2/x}$

Question

The derivative of function

$$f(x) = \sqrt{2x}+ \sqrt{2/x}$$

Here's what I did,

$$f(x) = \sqrt{2x}+ \sqrt{2/x} \\ = (2x)^{1\over2} + ({2\over x})^{1 \over 2}\\\\$$

$$f'(x)={1\over 2}(2x)^{-{1\over 2}}(2) + {1\over 2}({2 \over x})^{-{1 \over 2}}({2\over x^2}) \\=2x^{-{ 1\over 2}}+({1\over x})^{-{1\over 2}} ({2\over x^2})$$

I don't know what is going on!

Please help what I did wrong and what I should do

Thank you.

Or Can I see the steps how it's solved?

Answer 1

You have

\begin{align} f(x) &= \sqrt{2x} + \sqrt{\frac{2}{x}} \\ &= (2x)^{\frac{1}{2}} + (2x^{-1})^{\frac{1}{2}} \\ &= \frac{1}{2}\cdot 2 \cdot (2x)^{\frac{-1}{2}} + \frac{1}{2} \cdot -2x^{-2} \cdot (2x^{-1})^{\frac{-1}{2}} \\ &= \frac{1}{\sqrt{2x}} - \frac{1}{\sqrt{2} x^{\frac{3}{2}}} \end{align}